LeetCode #2813 — HARD

Maximum Elegance of a K-Length Subsequence

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed 2D integer array items of length n and an integer k.

items[i] = [profit_i, category_i], where profit_i and category_i denote the profit and category of the i^th item respectively.

Let's define the elegance of a subsequence of items as total_profit + distinct_categories², where total_profit is the sum of all profits in the subsequence, and distinct_categories is the number of distinct categories from all the categories in the selected subsequence.

Your task is to find the maximum elegance from all subsequences of size k in items.

Return an integer denoting the maximum elegance of a subsequence of items with size exactly k.

Note: A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements' relative order.

Example 1:

Input: items = [[3,2],[5,1],[10,1]], k = 2
Output: 17
Explanation: In this example, we have to select a subsequence of size 2.
We can select items[0] = [3,2] and items[2] = [10,1].
The total profit in this subsequence is 3 + 10 = 13, and the subsequence contains 2 distinct categories [2,1].
Hence, the elegance is 13 + 2² = 17, and we can show that it is the maximum achievable elegance.

Example 2:

Input: items = [[3,1],[3,1],[2,2],[5,3]], k = 3
Output: 19
Explanation: In this example, we have to select a subsequence of size 3. 
We can select items[0] = [3,1], items[2] = [2,2], and items[3] = [5,3]. 
The total profit in this subsequence is 3 + 2 + 5 = 10, and the subsequence contains 3 distinct categories [1,2,3]. 
Hence, the elegance is 10 + 3² = 19, and we can show that it is the maximum achievable elegance.

Example 3:

Input: items = [[1,1],[2,1],[3,1]], k = 3
Output: 7
Explanation: In this example, we have to select a subsequence of size 3. 
We should select all the items. 
The total profit will be 1 + 2 + 3 = 6, and the subsequence contains 1 distinct category [1]. 
Hence, the maximum elegance is 6 + 1² = 7.

Constraints:

1 <= items.length == n <= 10⁵
items[i].length == 2
items[i][0] == profit_i
items[i][1] == category_i
1 <= profit_i <= 10⁹
1 <= category_i <= n
1 <= k <= n

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed 2D integer array items of length n and an integer k. items[i] = [profiti, categoryi], where profiti and categoryi denote the profit and category of the ith item respectively. Let's define the elegance of a subsequence of items as total_profit + distinct_categories2, where total_profit is the sum of all profits in the subsequence, and distinct_categories is the number of distinct categories from all the categories in the selected subsequence. Your task is to find the maximum elegance from all subsequences of size k in items. Return an integer denoting the maximum elegance of a subsequence of items with size exactly k. Note: A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements' relative order.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Stack · Greedy

Example 1

[[3,2],[5,1],[10,1]]
2

Example 2

[[3,1],[3,1],[2,2],[5,3]]
3

Example 3

[[1,1],[2,1],[3,1]]
3

Related Problems

IPO (ipo)

Step 02

Core Insight

What unlocks the optimal approach

<div class="_1l1MA">Greedy algorithm.</div>
<div class="_1l1MA">Sort items in non-increasing order of profits.</div>
<div class="_1l1MA">Select the first <code>k</code> items (the top <code>k</code> most profitable items). Keep track of the items as the candidate set.</div>
<div class="_1l1MA">For the remaining <code>n - k</code> items sorted in non-increasing order of profits, try replacing an item in the candidate set using the current item.</div>
<div class="_1l1MA">The replacing item should add a new category to the candidate set and should remove the item with the minimum profit that occurs more than once in the candidate set.</div>

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2813: Maximum Elegance of a K-Length Subsequence
class Solution {
    public long findMaximumElegance(int[][] items, int k) {
        Arrays.sort(items, (a, b) -> b[0] - a[0]);
        int n = items.length;
        long tot = 0;
        Set<Integer> vis = new HashSet<>();
        Deque<Integer> dup = new ArrayDeque<>();
        for (int i = 0; i < k; ++i) {
            int p = items[i][0], c = items[i][1];
            tot += p;
            if (!vis.add(c)) {
                dup.push(p);
            }
        }
        long ans = tot + (long) vis.size() * vis.size();
        for (int i = k; i < n; ++i) {
            int p = items[i][0], c = items[i][1];
            if (vis.contains(c) || dup.isEmpty()) {
                continue;
            }
            vis.add(c);
            tot += p - dup.pop();
            ans = Math.max(ans, tot + (long) vis.size() * vis.size());
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2813: Maximum Elegance of a K-Length Subsequence
func findMaximumElegance(items [][]int, k int) int64 {
	sort.Slice(items, func(i, j int) bool { return items[i][0] > items[j][0] })
	tot := 0
	vis := map[int]bool{}
	dup := []int{}
	for _, item := range items[:k] {
		p, c := item[0], item[1]
		tot += p
		if vis[c] {
			dup = append(dup, p)
		} else {
			vis[c] = true
		}
	}
	ans := tot + len(vis)*len(vis)
	for _, item := range items[k:] {
		p, c := item[0], item[1]
		if vis[c] || len(dup) == 0 {
			continue
		}
		vis[c] = true
		tot += p - dup[len(dup)-1]
		dup = dup[:len(dup)-1]
		ans = max(ans, tot+len(vis)*len(vis))
	}
	return int64(ans)
}

# Accepted solution for LeetCode #2813: Maximum Elegance of a K-Length Subsequence
class Solution:
    def findMaximumElegance(self, items: List[List[int]], k: int) -> int:
        items.sort(key=lambda x: -x[0])
        tot = 0
        vis = set()
        dup = []
        for p, c in items[:k]:
            tot += p
            if c not in vis:
                vis.add(c)
            else:
                dup.append(p)
        ans = tot + len(vis) ** 2
        for p, c in items[k:]:
            if c in vis or not dup:
                continue
            vis.add(c)
            tot += p - dup.pop()
            ans = max(ans, tot + len(vis) ** 2)
        return ans

// Accepted solution for LeetCode #2813: Maximum Elegance of a K-Length Subsequence
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2813: Maximum Elegance of a K-Length Subsequence
// class Solution {
//     public long findMaximumElegance(int[][] items, int k) {
//         Arrays.sort(items, (a, b) -> b[0] - a[0]);
//         int n = items.length;
//         long tot = 0;
//         Set<Integer> vis = new HashSet<>();
//         Deque<Integer> dup = new ArrayDeque<>();
//         for (int i = 0; i < k; ++i) {
//             int p = items[i][0], c = items[i][1];
//             tot += p;
//             if (!vis.add(c)) {
//                 dup.push(p);
//             }
//         }
//         long ans = tot + (long) vis.size() * vis.size();
//         for (int i = k; i < n; ++i) {
//             int p = items[i][0], c = items[i][1];
//             if (vis.contains(c) || dup.isEmpty()) {
//                 continue;
//             }
//             vis.add(c);
//             tot += p - dup.pop();
//             ans = Math.max(ans, tot + (long) vis.size() * vis.size());
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2813: Maximum Elegance of a K-Length Subsequence
function findMaximumElegance(items: number[][], k: number): number {
    items.sort((a, b) => b[0] - a[0]);
    let tot = 0;
    const vis: Set<number> = new Set();
    const dup: number[] = [];
    for (const [p, c] of items.slice(0, k)) {
        tot += p;
        if (vis.has(c)) {
            dup.push(p);
        } else {
            vis.add(c);
        }
    }
    let ans = tot + vis.size ** 2;
    for (const [p, c] of items.slice(k)) {
        if (vis.has(c) || dup.length === 0) {
            continue;
        }
        tot += p - dup.pop()!;
        vis.add(c);
        ans = Math.max(ans, tot + vis.size ** 2);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.